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At a certain voltage applied to an `X`-rays tuve with aluminium anticathode the short-wave cut-off wavelength of the continuous `X`-rays spectum is equal to `0.50nm`. Will the `K` series of the characteristic specturm whose excitation potential is equal to `1.56kV` be also observed in this case? |
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Answer» Since the short wavelength cut off of the continuous specturm is `lambda_(0)=0.50nm` the voltage applied must be `V=(2 pi ħ )/(2 lambda_(0))=2.48kV` since this is greater than the excitation potential of the `K` series of the characteristic specturm (which is only `1.56kV`) the latter will be observed. |
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